The functions $f\left(x\right)$ and $g\left(x\right)$ are positive and continuous. If $f\left(x\right)$ is increasing and $g\left(x\right)$ is decreasing, then $\underset{0}{\overset{1}{\int }}f\left(x\right)\left[g\right(x)-g(1-x\left)\right]dx$

The correct option is A is always non-positive
Let $I=\underset{0}{\overset{1}{\int }}f\left(x\right)\left[g\right(x)-g(1-x\left)\right]dx\cdots \left(i\right)$
Using property:
$\underset{0}{\overset{a}{\int }}f\left(x\right)dx=\underset{0}{\overset{a}{\int }}f(a-x)dx$
$I=-\underset{0}{\overset{1}{\int }}f(1-x)\left[g\right(x)-g(1-x\left)\right]dx\cdots \left(ii\right)$
Adding equation $\left(i\right)$ and $\left(ii\right),$ we get
$2I=\underset{0}{\overset{1}{\int }}\left[f\right(x)-f(1-x\left)\right]\left[g\right(x)-g(1-x\left)\right]dx$
$\Rightarrow I=\frac{1}{2}\underset{0}{\overset{1}{\int }}\left[f\right(x)-f(1-x\left)\right]\left[g\right(x)-g(1-x\left)\right]dx\le 0$
$\because$ $f\left(x\right)$ is increasing function and $g\left(x\right)$ is decreasing function and
$f\left(x\right)-f(1-x)\le 0,g\left(x\right)-g(1-x)\ge 0$ for $x\in [0,\frac{1}{2}]$ and
$f\left(x\right)-f(1-x)\ge 0,g\left(x\right)-g(1-x)\le 0$ for $x\in [\frac{1}{2},1]$